I don't know how to create this tree in R - r

Question

I would like to maximize revenue by applying the better campaign at each hour.
I would like to create a tree that would help me choose the better campaign.
At the data below there's a record with the revenue per campaign per hour.
Looking at the data, I may see that campaign A is better between hours 1-12, and that campaign B is better between hours 13-24.
How do I create in R the tree that would tell me that?
hour campaign revenue
1 A 23
1 B 20
2 A 21
2 B 22
3 A 23
3 B 20
4 A 21
4 B 22
5 A 23
5 B 20
6 A 21
6 B 22
7 A 20
7 B 17
8 A 18
8 B 19
9 A 20
9 B 17
10 A 18
10 B 19
11 A 20
11 B 17
12 A 19
12 B 18
13 A 8
13 B 9
14 A 6
14 B 11
15 A 9
15 B 8
16 A 6
16 B 11
17 A 9
17 B 8
18 A 6
18 B 11
19 A 3
19 B 2
20 A 3
20 B 2
21 A 0
21 B 5
22 A 3
22 B 2
23 A 3
23 B 2
24 A 0
24 B 5

Answer 1

I'm not sure what kind of tree you are looking for exactly, but a linear model tree for revenue with regressor campaign and partitioning variable hour might be useful. Using lmtree() in package partykit you can fit a tree that starts out by fitting a linear model with two coefficients (intercept and campaign B effect) and then splits the data as long as there are significant instabilities in at least one of the coefficients:
library("partykit")
(tr <- lmtree(revenue ~ campaign | hour, data = d))
## Linear model tree
##
## Model formula:
## revenue ~ campaign | hour
##
## Fitted party:
## [1] root
## | [2] hour <= 12: n = 24
## | (Intercept) campaignB
## | 20.583333 -1.166667
## | [3] hour > 12: n = 24
## | (Intercept) campaignB
## | 4.666667 1.666667
##
## Number of inner nodes: 1
## Number of terminal nodes: 2
## Number of parameters per node: 2
## Objective function (residual sum of squares): 341.1667
In this (presumably artificial) data, this selects a single split at 12 hours and then has two terminal nodes: one with a negative campaign B effect (i.e., A is better) and one with a positive campaign B effect (i.e., B is better). The resulting plot(tr) yields:
This also brings out that the split is also driven by the change in revenue level and not only by the differing campaign effects (which are fairly small).
The underlying tree algorithm is called "Model-Based Recursive Partitioning" (MOB) and is also applicable to models other than linear regression. See the references in the manual and vignette for more details.
Another algorithm that might potentially be interesting is the QUINT (qualitative interaction trees) by Dusseldorp & Van Mechelen, available in the quint package.
For convenient replication of the example above: The d data frame can be recreated by
d <- read.table(textConnection("hour campaign revenue
1 A 23
1 B 20
2 A 21
2 B 22
3 A 23
3 B 20
4 A 21
4 B 22
5 A 23
5 B 20
6 A 21
6 B 22
7 A 20
7 B 17
8 A 18
8 B 19
9 A 20
9 B 17
10 A 18
10 B 19
11 A 20
11 B 17
12 A 19
12 B 18
13 A 8
13 B 9
14 A 6
14 B 11
15 A 9
15 B 8
16 A 6
16 B 11
17 A 9
17 B 8
18 A 6
18 B 11
19 A 3
19 B 2
20 A 3
20 B 2
21 A 0
21 B 5
22 A 3
22 B 2
23 A 3
23 B 2
24 A 0
24 B 5"), header = TRUE)

Answer 2

Would something like this work?
## create a sequence of hours from column 1 of the data
hr <- as.numeric(unique(data[,1]))
## Set up vectors to hold the A and B campaign "best" hours
A.hours=NULL
B.hours=NULL
## start at the lowest hour
i=1
while(i<=max(hr)) {
## create a subset of data from the current hour
sub.data <- data[matrix(which(data[,1]==hr[i])),]
## find the campaign with the highest revenue
best.camp <- sub.data[which(sub.data[,3]==max(sub.data[,3])),2]
if(best.camp=="A") {
A.hours <- c(A.hours,hr[i])
}
if(best.camp=="B") {
B.hours <- c(B.hours,hr[i])
}
i=i+1
}
The code indicates that during the A.hours (hours: 1 3 5 7 9 11 12 15 17 19 20 22 23), campaign A is more profitable.
However, during B.hours (hours: 2 4 6 8 10 13 14 16 18 21 24), campaign B is more profitable.